Problem: $h(t) = 7t^{2}-4(f(t))$ $f(n) = 3n^{2}-2n-7$ $ h(f(-1)) = {?} $
Explanation: First, let's solve for the value of the inner function, $f(-1)$ . Then we'll know what to plug into the outer function. $f(-1) = 3(-1)^{2}+(-2)(-1)-7$ $f(-1) = -2$ Now we know that $f(-1) = -2$ . Let's solve for $h(f(-1))$ , which is $h(-2)$ $h(-2) = 7(-2)^{2}-4(f(-2))$ To solve for the value of $h$ , we need to solve for the value of $f(-2)$ $f(-2) = 3(-2)^{2}+(-2)(-2)-7$ $f(-2) = 9$ That means $h(-2) = 7(-2)^{2}+(-4)(9)$ $h(-2) = -8$